This page will document useful transformations of elliptic integrals and relations among them, along with their derivations. These can be applied to simplifications of analytic formulae or numerical evaluations.

# Definitions of Integrals

The incomplete elliptic integrals of the first, second and third kinds in trigonometric form are

$F(φ|m) =∫0φ dφ 1-m sin2φ$

$E(φ|m) =∫0φ dφ 1-m sin2φ$

$Π(n; φ|m) =∫0φ dφ (1-n sin2φ) 1-m sin2φ$

with $0 in these defining integrals. The elliptic parameter m will be used throughout this presentation rather than the traditional elliptic modulus k for two reasons:

1) The resulting formulae are generally cleaner and simpler.

2) Computer algebra systems generally use the parameter rather than the modulus. Having formulae in the same form reduces unintended errors in application.

The elliptic parameter and modulus are related by $m={k}^{2}$. There is also traditionally a complementary elliptic modulus defined by ${{k}^{\prime }}^{2}=1-{k}^{2}$, and functions of this complementary modulus are often denoted with a prime.

While there is a variety of mathematical notation for the elliptic integrals, it appears to be common to distinguish those using the parameter with a vertical bar before that argument, while those using the modulus are denoted with a comma before that argument.

The transformation $x=sin\phi$ gives the incomplete elliptic integrals in Legendre normal form:

$F(φ|m) =∫0 sinφ dx (1-x2) (1-mx2 )$

$E(φ|m) =∫0 sinφ dx 1-mx2 1-x2$

$Π(n; φ|m) =∫0 sinφ dx (1-n x2) (1-x2) (1-mx2 )$

The angular variable is left as an argument on the left-hand sides to aid in avoiding errors in application.

Setting the angular argument equal to $π2$ defines the complete elliptic integrals:

$K(m) =∫0 π2 dφ 1-m sin2φ$

$E(m) =∫0 π2 dφ 1-m sin2φ$

$Π (n|m) =∫0 π2 dφ (1-n sin2φ) 1-m sin2φ$

# Complete Integrals of Negative Parameter

A simple angular substitution of the form

$φ=π2-θ sinφ=cosθ dφ=−dθ cosφ=sinθ$

interchanges sines and cosines while reversing the order of integration. This implies that the complete integrals can be equally defined using sines or cosines, since the integrals are otherwise identical after this substitution.

That fact can be used to evaluate the complete integrals for negative parameter quite easily:

$K(−m) =∫0 π2 dφ 1+m sin2φ =∫0 π2 dφ m+1-m cos2φ K(−m) =1m+1 K(m m+1)$

$E(−m) =∫0 π2 dφ 1+m sin2φ =∫0 π2 dφ m+1-m cos2φ E(−m) =m+1 E(m m+1)$

$Π(n| −m) =∫0 π2 dφ (1-n sin2φ) 1+m sin2φ Π(n| −m) =∫0 π2 dφ (1-n+n cos2φ) m+1-m cos2φ Π(n| −m) =1 (1-n) m+1 Π(n n-1| mm+1)$

# Reciprocal Modulus Transformations

With an argument of $1m$ the denominator of the complete integral of the first kind acquires a zero. Since the real part of the inverse sine function is always less than $π2$ , split the integral into two parts:

$K(1m) =∫0 sin−1 m dφ1 -1m sin2φ -i∫ sin−1 m π2 dφ1m sin2φ-1$

These two integrals can be converted to complete elliptic integrals with trigonometric substitutions that preserve the valid endpoint while scaling the other to a value appropriate for a complete integral. For the integral in the real part, the substitution that preserves the lower endpoint is

$sinφ=msinα cosφdφ =mcosα dα$

under which the integral becomes

$∫0 sin−1 m dφ1 -1m sin2φ =∫0 π2 mcosα dα 1-sin2α 1-msin2 α =mK(m)$

For the integral in the imaginary part, the substitution that preserves the upper endpoint is

$cosφ=1-m cosα sinφdφ =1-msinα dα$

under which the integral becomes

$∫ sin−1 m π2 dφ1m sin2φ-1 =∫0 π2 1-msinα dα 1-(1-m) cos2αm -1 1-(1-m) cos2α =mK (1-m)$

since the complete integrals can be defined with either sines or cosines as noted above. The reciprocal modulus transformation for the complete elliptic integral of the first kind is thus

$K(1m) =m[ K(m) -iK( 1-m)]$

This result technically only holds for $Imm>0$ : for $Imm<0$ the imaginary part of the result requires a change in sign.

For $m>1$ this relation can be used to separate the real and imaginary parts of the complete integral when written in the form

$K(m) =1m[ K(1m) -iK( 1-1m)]$

since then the arguments on the right-hand side are both within the defining range of convergence.

# Elliptic Nome on the Real Axis

The elliptic nome is defined by

$q(m) =exp[−π K(1-m) K(m)]$

For $0 the arguments of the elliptic integrals are within the defining range, so that both integrals as well as the nome are real.

For $m<0$ the arguments of both integrals are outside the defining range. The complete integral with negative parameter has been given above, and the integral with argument greater than one can be rewritten using the reciprocal modulus transformation. The ratio of the two integrals is

$K(1+m) K(−m) =m+1 K(m m+1) ×1m+1 [K(1 m+1) -iK(1-1 m+1)] =K(1 m+1) K(m m+1) -i$

so that the elliptic nome for real negative argument is

$q(−m) =−exp[−π K(1 m+1) K(m m+1) ]$

and since both arguments are now within the defining range, the nome is purely real.

For $m>1$ the arguments of both integrals are again outside the defining range but with their respective roles reversed. The ratio of the two integrals is now

$K(1-m) K(m) =K[ −(m-1)] K(m) K(1-m) K(m) =K(m-1 m)m ×m K(1m) -iK(1 -1m) K(1-m) K(m) =K(1 -1m) K(1m )2 +K(1 -1m )2 [K(1m) +iK(1 -1m)]$

The arguments on the right-hand side are now within the defining range, so that this ratio is explicitly separated into real and imaginary parts. The nome itself

$q(m) =exp[π K(1 -1m) K(1m )2 +K(1 -1m )2 [K(1m) +iK(1 -1m)]]$

is no longer entirely real in this domain.

Uploaded 2017.11.20 — Updated 2020.04.02 analyticphysics.com